Question

which of the following molecules exist in an ionized form at ph 9? use ka/kb table. there are 2 answers.

aniline, hypochlorus acid, phenol, ethylamine

Answer 1

Aniline has a pKb of 9.4 so its pKa is 14-9.4 = 4.6

Aniline + H⁺ aniliniumH⁺ below pH 4.6 right side is dominant

We know that Henderson-Hasselbalch equation

pKa = pH + log [HA] / [A^-]

4.6 = 9 + log [HA]/[A^-] To satisy this

log [HA]/[A^-] should be -4.4

so [HA]/[A^-] should be 3.98 x 10^-5 which means [HA] will be very small or negligible

pKa of hypochlorous acid is 7.4

HClO ClO^- + H⁺

pKa = pH + log [HA] / [A^-]

7.4 = 9 + log [HA]/[A^-] To satisy this

log [HA]/[A^-] should be -1.6

so [HA]/[A^-] should be 0.025 which means [HA] will be very small compared to [A^-] so it will be completely ionized

pKa of phenol is 10

phenol C₆H₅O^- + H⁺

pKa = pH + log [HA] / [A^-]

10 = 9 + log [HA]/[A^-] To satisy this

log [HA]/[A^-] should be 1

so [HA]/[A^-] should be 10 which means [A^-] will be 10% of [HA] so it will be partially ionised

pKb of ethylamine is 3.193 pKa of ethylamine is 14-3.193 = 10.8

ethylamine + H⁺ ethylammoniumH⁺

pKa = pH + log [HA] / [A^-]

10.8 = 9 + log [HA]/[A^-] To satisy this

log [HA]/[A^-] should be 1.8

so [HA]/[A^-] should be 64 which means there will be significant amount of HA and very small amount of A-

so the correct answer is hypochlorous acid and ethylamine will exist is ionised form at pH 9